Given a strictly increasing array arr of positive integers forming a sequence, return the length of the longest Fibonacci-like subsequence ofarr. If one does not exist, return 0.
A subsequence is derived from another sequence arr by deleting any number of elements (including none) from arr, without changing the order of the remaining elements. For example, [3, 5, 8] is a subsequence of [3, 4, 5, 6, 7, 8].
Example 1:
Input: arr = [1,2,3,4,5,6,7,8]
Output: 5
Explanation: The longest subsequence that is fibonacci-like: [1,2,3,5,8].
Example 2:
Input: arr = [1,3,7,11,12,14,18]
Output: 3
Explanation:The longest subsequence that is fibonacci-like: [1,11,12], [3,11,14] or [7,11,18].
Constraints:
3 <= arr.length <= 1000
1 <= arr[i] < arr[i + 1] <= 109
Solutions
Solution 1: Dynamic Programming
We define $f[i][j]$ as the length of the longest Fibonacci-like subsequence, with $\textit{arr}[i]$ as the last element and $\textit{arr}[j]$ as the second to last element. Initially, for any $i \in [0, n)$ and $j \in [0, i)$, we have $f[i][j] = 2$. All other elements are $0$.
We use a hash table $d$ to record the indices of each element in the array $\textit{arr}$.
Then, we can enumerate $\textit{arr}[i]$ and $\textit{arr}[j]$, where $i \in [2, n)$ and $j \in [1, i)$. Suppose the currently enumerated elements are $\textit{arr}[i]$ and $\textit{arr}[j]$, we can obtain $\textit{arr}[i] - \textit{arr}[j]$, denoted as $t$. If $t$ is in the array $\textit{arr}$, and the index $k$ of $t$ satisfies $k < j$, then we can get a Fibonacci-like subsequence with $\textit{arr}[j]$ and $\textit{arr}[i]$ as the last two elements, and its length is $f[i][j] = \max(f[i][j], f[j][k] + 1)$. We can continuously update the value of $f[i][j]$ in this way, and then update the answer.
After the enumeration ends, return the answer.
The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$, where $n$ is the length of the array $\textit{arr}$.